# partial fraction decomposition in integral

I have a question that asks to calculate the following integral:

$$\int_0^\infty {\frac{w\cdot \sin w}{4a^2+w^2}dw}$$

In the official solution they used partial fraction decomposition in order to later use Plancherel's identity:

$$\frac{w\cdot \sin w}{4a^2+w^2} =$$ $$\frac{\sin w}{w}\cdot\frac{w^2}{w^2+4a^2} =$$ $$\frac{\sin w}{w}(1-\frac{4a^2}{w^2+4a^2}) =$$ $$\frac{\sin w}{w} - \frac{\sin w}{w} \cdot \frac{4a^2}{w^2+4a^2}$$

And then used Plancherel's identity.

But I didn't understand how to expand to partial fraction and in particular I didn't understand this equation: $$\frac{\sin w}{w}\cdot\frac{w^2}{w^2+4a^2} = \frac{\sin w}{w}(1-\frac{4a^2}{w^2+4a^2})$$

Can you please explain how to expand the integrand into partial fraction?

• $$\frac{\sin w}{w}\Big(1-\frac{4a^2}{w^2+4a^2}\Big)= \frac{\sin w}{w}\Big(\frac{w^2+4a^2}{w^2+4a^2}-\frac{4a^2}{w^2+4a^2}\Big)=\frac{\sin w}{w}\Big(\frac{w^2}{w^2+4a^2}\Big)$$ – Axion004 Sep 15 at 1:46
• It is strange that the accepted answer is completely nonresponsive to the question asked: "Can you please explain how to expand the integrand into partial fraction?" – Eric Towers Sep 15 at 14:28

Consider the function:

$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w},\space a>0$$

Rewrite the integrand as follows:

$$\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}=\frac{w}{w}\cdot\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\frac{w^2}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\frac{w^2+4\cdot a^2-4\cdot a^2}{w^2+4\cdot a^2}=\frac{\sin(w)}{w}\cdot\left(\frac{w^2+4\cdot a^2}{w^2+4\cdot a^2}-\frac{4\cdot a^2}{w^2+4\cdot a^2}\right)=\frac{\sin(w)}{w}\cdot\left(1-\frac{4\cdot a^2}{w^2+4\cdot a^2}\right)=\frac{\sin(w)}{w}-\frac{\sin(w)}{w}\cdot\frac{4\cdot a^2}{w^2+4\cdot a^2}$$

Then:

$$I(a)=\int_0^\infty {\frac{\sin (w)}{w}\text{d}w}-\int_0^\infty {\frac{\sin(w)}{w}\cdot\frac{4\cdot a^2}{w^2+4\cdot a^2}\text{d}w}$$

The left-hand integral is known as a Dirichlet integral and it can be derived that it evaluates to $$\frac{\pi}{2}$$:

$$I(a)=\frac{\pi}{2}-\int_0^\infty {\frac{\sin(w)}{w}\cdot\frac{4\cdot a^2}{w^2+4\cdot a^2}\text{d}w}$$

Let $$w\mapsto 2\cdot a\cdot w$$:

$$I(a)=\frac{\pi}{2}-\int_0^\infty {\frac{\sin(2\cdot a\cdot w)}{2\cdot a\cdot w}\cdot\frac{4\cdot a^2}{(2\cdot a\cdot w)^2+4\cdot a^2}\cdot(2\cdot a\space\text{d}w)}=\frac{\pi}{2}-\int_0^\infty {\frac{\sin(2\cdot a\cdot w)}{ w}\cdot\frac{1}{w^2+1}\text{d}w}$$

Recognize that the integrand is a continuous and continuously differentiable function and differentiate with respect to $$a$$ under the integral sign:

$$I'(a)=\frac{\text{d}}{\text{d}w}\left[\frac{\pi}{2}-\int_0^\infty {\frac{\sin(2\cdot a\cdot w)}{ w}\cdot\frac{1}{w^2+1}\text{d}w}\right]=-\int_0^\infty {\frac{\partial}{\partial a}\frac{\sin(2\cdot a\cdot w)}{ w}\cdot\frac{1}{w^2+1}\text{d}w}=-2\cdot\int_0^\infty {\frac{w}{w}\cdot\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}=-2\cdot\int_0^\infty {\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}$$

Recognize that the integrand is a continuous and continuously differentiable function and differentiate with respect to $$a$$ under the integral sign:

$$I''(a)=-2\cdot\frac{\text{d}}{\text{d}a}\int_0^\infty {\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}=-2\cdot\int_0^\infty {\frac{\partial}{\partial a}\frac{\cos(2\cdot a\cdot w)}{w^2+1}\text{d}w}=4\cdot\int_0^\infty {\frac{w\cdot\sin(2\cdot a\cdot w)}{w^2+1}\text{d}w}$$

Consider the original expression for $$I(a)$$:

$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w}$$

Let $$w\mapsto 2\cdot a\cdot w$$:

$$I(a)=\int_0^\infty {\frac{2\cdot a\cdot w\cdot \sin(2\cdot a\cdot w)}{(2\cdot a\cdot w)^2+4\cdot a^2}\cdot(2\cdot a\space\text{d}w)}=\int_0^\infty {\frac{w\cdot \sin(2\cdot a\cdot w)}{w^2+1}\text{d}w}$$

Recognize that

$$I''(a)=4\cdot I(a)\Rightarrow I''(a)-4\cdot I(a)=0$$

Solving the differential equation yields

$$I(a) = \text{c}_{1}\cdot e^{2\cdot a} + \text{c}_{2}\cdot e^{-2\cdot a}$$

Differentiate with respect to $$a$$ on both sides:

$$I'(a) = 2\cdot\left(\text{c}_{1}\cdot e^{2\cdot a} - \text{c}_{2}\cdot e^{-2\cdot a}\right)$$

According to the closed form of $$I(a)$$, as $$a$$ approaches $$0$$, $$I(a\rightarrow 0)=\text{c}_{1}+\text{c}_{2}$$.

According to the integral form of $$I(a)$$, as $$a$$ approaches $$0$$, $$I(a\rightarrow 0)=\frac{\pi}{2}-\int_0^\infty {0\space\text{d}w}=\frac{\pi}{2}$$.

According to the closed form of $$I'(a)$$, as $$a$$ approaches $$0$$, $$I'(a\rightarrow 0)=2\cdot(\text{c}_{1}-\text{c}_{2})$$.

According to the integral form of $$I'(a)$$, as $$a$$ approaches $$0$$, $$I'(a\rightarrow 0)=-2\cdot\int_0^\infty {\frac{1}{w^2+1}\text{d}w}=-2\cdot\frac{\pi}{2}=-\pi$$.

It can be derived that $$\text{c}_{1}=0$$ and $$\text{c}_{2}=\frac{\pi}{2}$$.

Then,

$$I(a)=\int_0^\infty {\frac{w\cdot \sin (w)}{w^2+4\cdot a^2}\text{d}w}=\frac{\pi}{2}\cdot e^{-2\cdot a},\space a>0$$

There is no (nontrivial) partial fraction decomposition in the manipulations you described. The denominator is irreducible (over the reals), so that method normally is thwarted (but see below)

Consider $$\frac{a}{a+b} = \frac{a+b-b}{a+b} = 1- \frac{b}{a+b} \text{.}$$

There is the possibility of a partial fractions decomposition, PFD, over the complex numbers. To perform any PFD, you must be able to combine fractions under the integral to a single fraction with a factored denominator. Since $$4a^2 + w^2 = (2a+\mathrm{i}w)(2a-\mathrm{i}w) \text{,}$$ both of which are linear in the variable of integration, $$w$$, so we would look for $$u$$ and $$v$$ such that $$\frac{u}{2a+\mathrm{i}w} + \frac{v}{2a-\mathrm{i}w} \text{.}$$ Multiplying out and equating likes, we have $$u = \mathrm{i}/2$$ and $$v = -\mathrm{i}/2$$, so $$\frac{w}{4a^2 + w^2} = \frac{\mathrm{i}/2}{2a+\mathrm{i}w} - \frac{\mathrm{i}/2}{2a-\mathrm{i}w} \text{.}$$ Of course, everyone knows $$\int_0^\infty \; \frac{\mathrm{i}/2}{2a+\mathrm{i}w} \,\mathrm{d}w = \frac{1}{4} ((\pi +2 \mathrm{i} \, \mathrm{Shi}(2 a)) \cosh (2 a)-2 \mathrm{i} \, \mathrm{Ci}(-2 \mathrm{i} a) \sinh (2 a)) \text{,}$$ where $$\mathrm{Shi}$$ is the hyperbolic sine integral, $$\cosh$$ is the hyperbolic cosine, $$\mathrm{Ci}$$ is the cosine integral, and sinh is the hyperbolic sine. Or, to summarize, this is not the way to go for this problem.

$$1=\frac{w^2}{w^2+4a^2}+\frac{4a^2}{w^2+4a^2}$$